We will learn how to find the general values of inverse trigonometric functions in different types of problems.
1. Find the general values of sin\(^{-1}\) (- √3/2)
Solution:
Let, sin\(^{-1}\) (- √3/2) = θ
Therefore, sin θ = - √3/2
⇒ sin θ = - sin (π/3)
⇒ sin θ = (- π/3)
Therefore, the general value of sin\(^{-1}\) (- √3/2) = θ = nπ - (- 1)\(^{n}\) π/3, where, n = 0 or any integer.
2.
Find the general values of cot\(^{-1}\) (- 1)
Solution:
Let, cot\(^{-1}\) (- 1) = θ
Therefore, cot θ = - 1
⇒ cot θ = cot (- π/4)
Therefore, the general value of cot\(^{-1}\) (- 1) = θ = nπ - π/4, where, n = 0 or any integer.
3. Find the general values of cos\(^{-1}\) (1/2)
Solution:
Let, cos\(^{-1}\) 1/2 = θ
Therefore, cos θ = 1/2
⇒ cos θ = cos (π/3)
Therefore, the general value of cos\(^{-1}\) (1/2) = θ = 2nπ ± π/3, where, n = 0 or any integer.
4. Find the general values of sec\(^{-1}\) (- 2)
Solution:
Let, sec\(^{-1}\) (- 2) = θ
Therefore, sec θ = - 2
⇒ sec θ = - sec (π/3)
⇒ sec θ = sec (π - π/3)
⇒ sec θ = sec (2π/3)
Therefore, the general value of sec\(^{-1}\) (- 2) = θ = 2nπ ± 2π/3, where, n = 0 or any integer.
5. Find the general values of csc\(^{-1}\) (√2)
Solution:
Let, csc\(^{-1}\) (√2) = θ
Therefore, csc θ = √2 .
⇒csc θ = csc (π/4)
Therefore, the general value of csc\(^{-1}\) (√2 ) = θ = nπ + (- 1)\(^{n}\) π/4, where, n = 0 or any integer.
6. Find the general values of tan\(^{-1}\) (√3)
Solution:
Let, tan\(^{-1}\) (√3) = θ
Therefore, tan θ = √3
⇒ tan θ = tan (π/3)
Therefore, the general value of tan\(^{-1}\) (√3) = θ = nπ + π/3 where, n = 0 or any integer.
● Inverse Trigonometric Functions
11 and 12 Grade Math
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